Benders Decomposition for Solving Two-stage Stochastic Optimization Models

Transcription

1 Benders Decomposition for Solving Two-stage Stochastic Optimization Models IMA New Directions Short Course on Mathematical Optimization Jim Luedtke Department of Industrial and Systems Engineering University of Wisconsin-Madison August 9, 2016 Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 1 / 33

2 Review Two-Stage Stochastic LP with Recourse where for s = 1,..., S How to solve this problem? z SP = min c x + S s=1 p sq s (x) s.t. Ax b x R n 1 + Q s (x) def = Q(x, ξ s ) = min q s y s.t. W s y = h s T s x y R n 2 + Yesterday: Deterministic equivalent form Large-scale LP Today: Benders decomposition Exploit structure Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 3 / 33

3 Common Assumption Relatively Complete Recourse For every feasible first-stage feasible solution (Ax b, x R n 1 + ) and every scenario s there exists a feasible recourse decision y s (W s y s = h s T s x, y s R n 2 + ). Simplifies Benders decomposition algorithm We ll first present the algorithm under this assumption Then relax the assumption Assumption is very important when using sample average approximation Relatively complete recourse should hold for all possible realizations of random outcomes Otherwise, solution to SAA problem may not even be feasible to original problem Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 4 / 33

4 Scenario Value Function Primal Dual Q s (x) = min y q s y s.t. W s y = h s T s x max π (h s T s x) π s.t. π W s q s y R n 2 + Assumption Π s := {π : π W s q s }. If this assumption is violated: For every x, second-stage primal problem is either infeasible or unbounded Stochastic LP is either infeasible or unbounded Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 5 / 33

5 Scenario Value Function Primal Dual Q s (x) = min y q s y s.t. W s y = h s T s x max π (h s T s x) π s.t. π W s q s y R n 2 + Assumption Π s := {π : π W s q s }. Relatively complete recourse Primal has a feasible solution Strong duality applies: Primal and dual each have an optimal solution Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 5 / 33

6 Using Strong Duality Assuming Π s and relatively complete recourse: Q s (x) = min y {q s y : W s y = h s T s x, y R n 2 + } = max π {π (h s T s x) : π W s q s } = max{(π s ) (h s T s x) : π s XP(Π s )} where XP(Π s ) is the finite set of extreme points of Π s. Structure of Q s ( ) Q s ( ) is a piecewise-linear convex function. Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 6 / 33

7 Benders Reformulation z SP = min x Observe: c x + S s=1 p sq s (x) = min x,θ c x + S s=1 p sθ s s.t. Ax b, x R n 1 + s.t. Ax b, x R n 1 + θ s Q s (x), s = 1,..., S θ s Q s (x) θ s max π {π (h s T s x) : π W s q s } θ s max{(π s ) (h s T s x) : π s XP(Π s )} θ s (π s ) (h s T s x), π s XP(Π s ) Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 7 / 33

8 Benders Reformulation z SP = min x,θ c x + S s=1 p sθ s s.t. Ax b, x R n 1 + θ s (π s ) (h s T s x), s = 1,..., S, π s XP(Π s ) Explicit linear program formulation Many fewer variables than deterministic equivalent form (just 1 per scenario) Potentially HUGE number of constraints Theoretically solvable by ellipsoid algorithm Practically solvable by a cutting plane algorithm Benders decomposition/l-shaped algorithm Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 8 / 33

9 Benders Decomposition (L-Shaped Method) At iteration t of Benders decomposition, a Master Problem (MP) is solved: (MP) t : z t := min θ,x c x + S p s θ s s=1 where V s,t XP(Π s ) for s = 1,..., S s.t. Ax b, x R n 1 + θ s (π s ) (h s T s x), s = 1,..., S, π s V s,t V s,t t, so constraints of (MP) t are a (small) subset of constraints of Benders reformulation z t z SP Given an optimal solution (ˆx t, ˆθ t ) to (MP) t, we must determine if any of the excluded constraints are violated Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 9 / 33

10 Benders Subproblems Let (ˆx t, ˆθ t ) be an optimal solution to (MP) t : For each s, are the following constraints all satisfied? ˆθ t s (π s ) (h s T sˆx t ), π s XP(Π s ) True if and only if: ˆθ t s max{π (h s T sˆx t ) : π Π s } = Q s (ˆx t ) Thus, for each s, we solve the second-stage subproblem: Q s (ˆx t ) = max π {π (h s T sˆx t ) : π W s = q s } = min y {q s y : W s y = h s T s x, y R n 2 + } Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 10 / 33

11 Benders Subproblems For each s, we solve the second-stage subproblem: Q s (ˆx t ) = max π {π (h s T sˆx t ) : π W s = q s } If ˆθ t s Q s (ˆx t ): Do nothing! = min y {q s y : W s y = h s T s x, y R n 2 + } All the constraints for scenario s are satisfied If ˆθ t s < Q s (ˆx t ): Let ˆπ s be an extreme point optimal dual solution The constraint θ s (ˆπ s ) (h s T s x) is violated by (ˆx t, ˆθ t ) Add to the master problem: V s V s {ˆπ s } Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 11 / 33

12 Upper bounds At each iteration, master problem objective value provides a lower bound on z SP Because ˆx t is a feasible solution, we also obtain an upper bound after having solved the scenario subproblems: z SP c ˆx t + S p s Q s (ˆx t ) Thus, we can terminate the algorithm when these bounds are equal, or close enough (e.g., within ɛ) s=1 Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 12 / 33

13 Recap: Benders Decomposition Assume for simplicity: {x R n 1 + : Ax b} is bounded Initialization: Let ˆx 0 be an optimal solution to min{c x : Ax b, x R n 1 + } For each scenario s, solve scenario subproblem Q s (ˆx 0 ), let ˆπ s be an optimal dual solution, and set V s = {ˆπ s }. For t = 1, 2, Solve Master Problem to obtain solution (ˆx t, ˆθ t ) with objective value ẑ t 2. For each s = 1,..., S: Solve scenario subproblem to evaluate Q s (ˆx t ), and let ˆπ s be an optimal dual solution If ˆθ t s < Q s (ˆx t ): Set V s V s {ˆπ s } 3. If ẑ t c ˆx t + S s=1 p sq s (ˆx t ) ɛ: Break. Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 13 / 33

14 Convergence The algorithm trivially terminates finitely Only finitely many extreme point dual solutions If no new cuts added, lower and upper bounds will be equal But no guarantee on rate of convgerence (similar to simplex) In practice, often converges pretty fast, but occasionally disastrous Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 14 / 33

15 Implementation Notes Must use a cut violation tolerance, e.g., δ 10e 6 : Numerically, conclude ˆθ t s < Q s (ˆx t ) only if ˆθ t s < Q s (ˆx t ) δ δ should match the feasibility tolerance of LP solver Otherwise, you may add a cut that LP solver does not think is violated infinite loop Not necessary to solve all scenario subproblems in every iteration But only obtain an upper bound (and hence can consider terminating) when all are solved Focus on useful scenarios Scenario subproblems can be solved in parallel Parallel scalability eventually limited by master problem Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 15 / 33

16 Example A First Example subject to min x 1 + x 2 ω 1 x 1 + x 2 7 ω 2 x 1 + x 2 4 x 1 0 x 2 0 ω = (ω 1, ω 2 ) Ω = {(1, 1/3), (5/2, 2/3), (4, 1)} Each outcome has p s = 1 3 Huh? This problem doesn t make sense! Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 16 / 33

17 Example Recourse Formulation where min x 1 + x s=1 p sq s (x) s.t. x 1, x 2 0 Q s (x) = min y 1 + y 2 ω 1s x 1 + x 2 + y 1 7 ω 2s x 1 + x 2 + y 2 4 y 1, y 2 0 (ω 1s, ω 2s ) {(1, 1/3), (5/2, 2/3), (4, 1)} Each outcome has p s = 1 3 Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 17 / 33

18 Handling Infeasible Subproblems Without Relatively Complete Recourse Recall the second-stage scenario subproblem: Q s (x) = min y {q s y : W s y = h s T s x, y R n 2 + } Without relatively complete recourse, this problem might be infeasible for some x: Yields implicit constraints on x Let C s = {x R n 1 : y R n 2 + s.t. W sy = h s T s x} Need to make these implicit constraints explicit in two-stage formulation: min c x + S s=1 p sq s (x) s.t. Ax b, x R n 1 + x C s, s = 1,..., S Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 19 / 33

19 Handling Infeasible Subproblems Characterizing implicit constraints For what x is this problem feasible? Recall the dual: Q s (x) = min y {q s y : W s y = h s T s x, y R n 2 + } max π {π (h s T s x) : π W s q s } Dual assumed to be feasible: Primal feasible Dual bounded Dual bounded (r s ) (h s T s x) 0 for every extreme ray r s XR(Π s ) of dual feasible region Π s Thus, subproblem for scenario s is feasible if and only if x C s := {x R n 1 : (r s ) (h s T s x) 0, r s XR(Π s )} Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 20 / 33

20 Handling Infeasible Subproblems Modified Benders Reformulation min x,θ c x + S s=1 p sθ s s.t. Ax b, x R n 1 + θ s (π s ) (h s T s x), s = 1,..., S, π s XP(Π s ) (r s ) (h s T s x) 0, s = 1,..., S, r s XR(Π s ) Again can solve with a cutting plane algorithm Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 21 / 33

21 Handling Infeasible Subproblems Solving Modified Benders Reformulation Master problem: Replace full set of extreme points XP(Π s ) with a subset V s Replace full set of extreme rays XR(Π s ) with a subset R s Given a master solution (ˆx t, ˆθ t ) solve each scenario s subproblem: If subproblem s feasible and ˆθ t s < Q s (ˆx t ): Add optimality cut : V s V s {ˆπ s } If subproblem s is infeasible: Simplex algorithm yields a dual extreme ray ˆr s with (ˆr s ) (h s T sˆx t ) > 0 Add feasibility cut : R s R s {ˆr s } Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 22 / 33

22 Single-cut Benders Single-cut vs. Multi-cut Benders Benders algorithm we have seen is referred to as multi-cut version Cuts are used to approximate value function of each scenario Many cuts may be added in each iteration Alternative: Single-cut implementation Define Q(x) = S s=1 p sq s (x) Basis of Benders reformulation: A single variable Θ Any scenario subproblem infeasible Q(x) not defined: Enforce all feasibility cuts as in multi-cut Benders min x,θ c x + Θ s.t. Ax b, x R n 1 + (r s ) (h s T s x) 0, s = 1,..., S, r s XR(Π s ) Θ Q(x) Need explicit reformulation of last constraint... Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 24 / 33

23 Single-cut Benders Single-cut Benders decomposition Assuming all scenario subproblems are feasible for some x: Q(x) = = = = S p s Q s (x) s=1 S s=1 S s=1 p s min y {q s y : W s y = h s T s x, y R n 2 + } p s max π {π (h s T s x) : π W s = q s } S p s max{(π s ) (h s T s x) : π s XP(Π s )} s=1 Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 25 / 33

24 Single-cut Benders Single-cut Benders decomposition Assuming all scenario subproblems are feasible for some x: S Q(x) = p s max{(π s ) (h s T s x) : π s XP(Π s )} s=1 Therefore Θ Q(x) if and only if: S Θ p s (π s ) (h s T s x), (π 1,..., π S ) XP(Π 1 ) XP(Π S ) s=1 HUGE number of constraints But for cutting plane algorithm only need to be able to efficiently find a violated constraint Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 26 / 33

25 Single-cut Benders Single-cut Benders master problem Master problem after adding t optimality cuts min x,θ c x + Θ s.t. Ax b, x R n 1 + Θ d k c k x, (r s ) (h s T s x) 0, s = 1,..., S, r s R s k = 1,..., t The constraints Θ d k c k x, k = 1,..., t are just a compact way of writing a subset of these constraints: Θ S p s (π s ) (h s T s x), (π 1,..., π S ) XP(Π 1 ) XP(Π S ) s=1 Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 27 / 33

26 Single-cut Benders Single-cut Benders master problem Let (ˆx, ˆΘ) be a master problem optimal solution Solve all scenario subproblems for this x If any one of them is infeasible: Add feasibility cut(s) Else if ˆΘ S s=1 p sq s (ˆx): Solution is feasible, hence optimal Else: Let ˆπ s be optimal extreme point dual solution, s = 1,..., S Add following cut which is violated by (ˆx, ˆΘ): Θ S p s (ˆπ s ) (h s T s x) := d t+1 c t+1x s=1 where d t+1 = S s=1 p s(ˆπ s ) h s and c t+1 = S s=1 (ˆπs ) T s Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 28 / 33

27 Single-cut Benders Single-cut vs. Multi-cut Benders Single-cut Benders Fewer variables and optimality cuts Master problem typically solves faster Must solve all subproblems to generate an optimality cut Multi-cut Benders Many cuts added per iteration Typically converges in many fewer interations Master problem may become large and slow to solve Do not need to solve all subproblems every iteration Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 29 / 33

28 Cut Selection Cut Selection (in Multi-cut Benders) Given (ˆx, ˆθ) optimal to master problem, key step is finding a violated cut: Optimality cut: Dual solution ˆπ s with ˆθ s < (ˆπ s ) (h s T sˆx) Feasibility cut: Dual extreme ray ˆr s with (ˆr s ) (h s T sˆx) > 0 Benders method provides a recipe for finding some violated constraint whenever one exists But, there may be many different violated constraints Good choice can lead to faster convergence Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 31 / 33

29 Cut Selection Alternative Cut Generation Problem Proposed by Fischetti, Salvagnin, and Zanette (2010): Theorem ˆv s = max π (h s T sˆx) ˆθ s π 0 s.t. π W s q s π 0 π 1 + π 0 = 1 ˆv s > 0 if and only if either there exists ˆπ s XP(Π s ) with or there exists ˆr s XR(Π s ) with ˆθ s < (ˆπ s ) (h s T sˆx) (ˆr s ) (h s T sˆx) > 0. Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 32 / 33

30 Cut Selection Alternative Cut Generation Problem Conclusion from Theorem: ˆv s = max π (h s T sˆx) ˆθ s π 0 s.t. π W s q s π 0 π 1 + π 0 = 1 We can solve this problem for each scenario to generate either feasibility or optimality cuts But why? This problem searches for a deepest cut in some sense Empirical evidence suggests using cuts found this way can dramatically reduce solution times Jim Luedtke (UW-Madison) Benders Decomposition Lecture Notes 33 / 33